https://file.aiquickdraw.com/mathbot/file/1726259031237ucdmfi6d.jpg

Let's break down the problem from the image:

1. The function that models the bacteria population in one cubic centimeter of blood is given by:
   \[
   P(t) = 85t(0.89^t)
   \]
   where \( P(t) \) represents the bacteria population, and \( t \) is the time in days since the person became ill.

### Questions:

1. **When is the bacteria population at a maximum?**
   - To find this, we need to determine when the derivative of \( P(t) \) equals 0, which indicates the critical point.

2. **What is the maximum population?**
   - After determining the day at which the population is maximized (from Question 1), substitute that value of \( t \) into the function \( P(t) \) to find the population.

3. **Rate of change of the population at \( t = 5 \):**
   - This requires calculating the derivative of \( P(t) \) at \( t = 5 \).

I will solve these questions step by step. Let's first calculate the derivative of \( P(t) \), solve for the critical points to find when the population is at its maximum, and then calculate the rate of change at day 5.The derivative of the population function \( P(t) = 85t(0.89^t) \) is:

\[
P'(t) = -9.91 \cdot 0.89^t \cdot t + 85 \cdot 0.89^t
\]

Solving for when \( P'(t) = 0 \), we find the critical point occurs at approximately \( t = 8.58 \) days. This means that the bacteria population reaches its maximum around day 9 (rounded to the nearest day, as requested).

Now, let's calculate the population at this critical point to determine the maximum population.It seems like I can't do more advanced data analysis right now. Please try again later.

Here are the steps you can follow to complete the rest of the problem manually:

1. **Maximum Population**:
   - We found that the maximum occurs around day 8.58. To find the population at this time, substitute \( t = 8.58 \) into the function:
     \[
     P(8.58) = 85(8.58)(0.89^{8.58})
     \]
     Calculate this to find the maximum population.

2. **Rate of Change at Day 5**:
   - To estimate the rate of change at day 5, use the derivative we found:
     \[
     P'(t) = -9.91 \cdot 0.89^t \cdot t + 85 \cdot 0.89^t
     \]
     Substitute \( t = 5 \) into this equation to find the rate of change at day 5.

Let me know if you'd like further clarification or help with specific calculations!

### Relative questions:
1. What does the derivative tell you about the population behavior?
2. Why is the population decreasing after the maximum point?
3. How does exponential decay influence the bacteria population over time?
4. Can the bacteria population ever reach zero? Why or why not?
5. How would changing the constant 85 in the equation affect the graph?

#### Tip:
Always double-check the units in your final answer, especially when interpreting rates or population values.

The population of bacteria in one cubic centimeter of the blood of a sick person has been modeled by the function P(t) = 85t(0.89^t) where t is the time, in days, since the person became ill. To the nearest day, when is the bacteria population at a maximum? What is the maximum population? Round your answer to one decimal place. Estimate how fast the population is changing 5 days after the onset of illness. Round your answer to two decimal places.

Solve the problem of finding the maximum population of bacteria and estimate the rate of change in the population 5 days after the onset of illness using calculus. The function P(t) = 85t(0.89^t) models exponential decay.

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